\newproblem{lay:7_4_21}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.21}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Jan. 20th, 2014} \\}{}

  % Problem statement
	Show that if $P$ is an orthogonal $m\times m$ matrix, then $PA$ has the same singular values as $A$.
}{
   % Solution
	We know that the singular values of a matrix $A$ are the square root of the eigenvalues of the matrix $A^TA$. The singular values of $PA$ will be the square root of the eigenvalues of the matrix
\begin{equation}
	(PA)^T(PA)=A^TP^TPA=A^TA
\end{equation}
where we have made use of the fact that $P$ is orthogonal and, consequently, $P^TP=I$.
}
\useproblem{lay:7_4_21}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

